Evaluate the integral inside domain V, where V is bounded by the planes

[Cloudless]

1. Evaluate the integral

∫_VxdV

inside domain V, where V is bounded by the planes x=0, y=x, z=0, and the surface x²+y²+z²=1


Answer given: 1/8 - √2/16 (which is NOT what I got.. )
2. The attempt at a solution

Ok, it's a triple integral, I know this.

∫dx runs from 0 to 1

∫dy runs from -√1-x² to x

∫dz runs from 0 to √1-x²-y²

So the order of integrals go ∫∫∫dz dy dx

Yeah... it's not working. I keep getting stuck with things like x(1-y²-x²)^1/2 which I can't integrate onwards...

 

[tiny-tim]

Hi Cloudless!

∫_VxdV

inside domain V, where V is bounded by the planes x=0, y=x, z=0, and the surface x²+y²+z²=1

this is a vertically sliced wedge from θ = 45° to 90°, so it would be much easier to use spherical coordinates

of course, you can use x y z, but …

∫dx runs from 0 to 1

no

∫dy runs from -√1-x² to x

no

 

[HallsofIvy]

In the xy-plane, z= 0 so $x^2+ y^2+ z^2= 1$ projects to the quarter circle $x^2+ y^2= 1$. For each x, y ranges for 0 up to $y= \sqrt{1- x^2}$, NOT $y= -\sqrt{1- x^2}$ because the bounding surface is above the xy-plane.

 

[Cloudless]

I redid it in spherical coordinates. Much easier, but still missing a pi/2. I think the conclusion is the answer key was wrong.

In the xy-plane, z= 0 so $x^2+ y^2+ z^2= 1$ projects to the quarter circle $x^2+ y^2= 1$. For each x, y ranges for 0 up to $y= \sqrt{1- x^2}$, NOT $y= -\sqrt{1- x^2}$ because the bounding surface is above the xy-plane.

The question was ambiguous in its bounded area. I thought it was from negative pi/2 to pi/4.

 

[tiny-tim]

I redid it in spherical coordinates. Much easier, but still missing a pi/2. I think the conclusion is the answer key was wrong.

If you want us to check your answer, you'll have to show us your calculations.